Problem 8-28 The exponential distribution applies to lifetimes of a certain component. Its failure rate is unknown. Find the probability that the component will survive past 5 years assuming:

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Problem 8-28 The exponential distribution applies to lifetimes of a certain component. Its failure rate is unknown. Find the probability that the component will survive past 5 years assuming:

(a) lambda=.5

Pr=

(b) lambda=0.9

Pr=

(c) lambda=1.1

Pr=

answer : a) P( X> 5 ) = 0.0821

              b) P( X > 5 ) = 0.0111

              c) P ( X > 5 ) = 0.00409

Step-by-step explanation:

A) P( X > 5 ) = exp ( - λ * x )

   = ≈ 0.082

B)  P( X > 5 ) = exp ( - λ * x )  

  = ≈ 0.0111

C) P( X > 5 ) = exp ( - λ * x )  

 =  ≈ 0.00409

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

1 x=22.52

2: 65/39=604+39x

65x-39x=604

26x=604

x=604/26

×=23.23